Q1) Let $n=3$ and let $$\omega =F_1dx_1+F_2dx_2+f_3dx_3.$$ Then $$d \omega =K_1 dx_2\wedge dx_3+K_2 dx_3\wedge dx_1+K_3 dx_1\wedge dx_2,$$ where $(K_1,K_2,K_3)=\operatorname{Curl}(F)$. Therefore $d\omega $ gives again the operation of the curling.
What does it mean ? I don't really understand. Because the curling is indeed a differential operator but is not a differential form.
Q2) Same if $\omega $ is a $p-$form we have for $f\in \mathcal C^1$ that $$d(f\omega )=df\wedge \omega +f(d\omega ),$$ which is equivalent to $$\nabla (fg)=f\nabla g+g\nabla f$$ if $g\in \mathcal C^1$. Again, le gradient is a differential operator, what is the connection with differential form ?
Well, you've already written the answer yourself: if you take the coefficients of the differential forms $\omega$ and $d\omega$ and build vectors $F$ and $K$ out of them, then $K$ is the curl of $F$.